In the Neveu-Schwarz sector, the worldsheet fermions can be expanded as

$$ \psi^I(\tau,\sigma) \sim \sum\limits_{r\in Z+1/2}b_r^Ie^{-ir(\tau-\sigma)} $$

and the total mass squared operator can then be written as

$$ M^2 = \frac{1}{\alpha'}\left( \frac{1}{2} \sum\limits_{p\neq 0} \alpha_{-p}^I\alpha_p^I + \frac{1}{2}\sum\limits_{r\in Z+1/2} r \, b_{-r}^I b_r^I \right) $$

The first sum gives the contribution of the bosons, the second one the contributions of the fermions.

Why are the summands in in the fermionic part multiplied by $r$, how does this factor come in mathematically? Does this have something to do with the Pauli exclusion principle?

The same thing issue appears with the fermionic part mass operator in the Ramond sector, where I dont understand it either ...

  • 1
    $\begingroup$ @DImension10AbhimanyuPS if you find the answer, you could post it here :-). When trying to look at the lecture notes, the nasty firewall we have at work (that should not exist as two young bright colleagues of Lumo nicely proved ...) intervenes, so I'll look at it at home. $\endgroup$ – Dilaton Oct 28 '13 at 9:58

If one solves the field equations for the bosonic field, with the Newmann/Dirchilet/Closed String Boundary conditions, one can see that the mode expansion is something like:

$$X^\mu=...+i\sqrt{2\alpha'} \sum_{n\neq0 }^{ } \frac{\alpha^\mu}{n}\exp\left(in\sigma^0\right)\cos\left(n\sigma^1\right)$$

On the other hand, the fermionic field mode expansion goes like:

$$\psi^\mu_\pm = \frac1{\sqrt2}\sum_{n\in\mathbb Z \ \mathrm{or} \ \mathbb{Z}+\frac12 }b_r^\mu \exp\left(-ir\sigma^\pm\right) $$

Notice that there is a missing factor of $\frac1r$ in the second equation.

$[N,b_{-r}^\mu]=rb_{-r}^\mu$ and the conclusion follows.

  • $\begingroup$ Gosh, I'll have to check if the factor 1/r and maybe a factor r in the commutator is missing in my Zweibach book too or if it is just missing in my question ... Your answer will then solve the issue, thanks :-). In fact, thnking about it I should have seen that it is missing in the fermion expansion myself, darn ...! $\endgroup$ – Dilaton Oct 28 '13 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.